Answer
$\lim\limits_{h\to0}\frac{(3+h)^{-1}-3^{-1}}{h}=\frac{-1}{9}$
Work Step by Step
$\lim\limits_{h\to0}\frac{(3+h)^{-1}-3^{-1}}{h}$
$=\lim\limits_{h\to0}\frac{\frac{1}{3+h}-\frac{1}{3}}{h}$
$=\lim\limits_{h\to0}\frac{\frac{3-(3+h)}{3(3+h)}}{h}$
$=\lim\limits_{h\to0}\frac{\frac{-h}{3(3+h)}}{h}$
$=\lim\limits_{h\to0}\frac{-h}{3h(3+h)}$
$=\lim\limits_{h\to0}\frac{-1}{3(3+h)}$ (divide both numerator by denominator by $h$)
$=\frac{-1}{3(3+0)}$
$=\frac{-1}{9}$
